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2 years ago

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Technician A that shielding gas nozzles may have different shapes. Technician B says that gelding gas nozzles is attached to the

shielding gas cylinder. Who is right?

1 answer:

Lilit [14]2 years ago

5 0

B is correct! in this senecio

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The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought

harkovskaia [24]

Answer:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, $M_s$ = 600 g

mass of dry soil sample, $M_d$ = 200 g

mass of water content, $M_w$ = $M_s$ - $M_d$ = 600 g - 200 g = 400 g

The water content is determined as;

$M_w(\%) = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%) = \frac{600-200}{600} *100 \% \\\\M_w(\%) = 66.67 \%$

Therefore, the water of the saturated clayed soil is 66.67 %.

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2 years ago

Prelest! Introduction to Engineering and Technology 1 Select the correct answer. What technological invention allowed for the pr

satela [25.4K]

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2 years ago

A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn

topjm [15]

Answer:

k = 1.91 × 10^-5 N m rad^-1

Workings and Solution to this question can be viewed in the screenshot below:

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2 years ago

ANSWER FAST PLEASE!!! WILL MARK BRAINLIEST!!!!!!

Phoenix [80]

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For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

2 years ago